Quadratic Equations | Revision Notes
- An equation of the form ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0, is called a quadratic equation in x.
- A real number α is called a root of the quardratic equation ax2 + bx + c = 0, a ≠ 0, if aα2 + bα + c =0.
Any quardratic equation can have at most two roots.
Note : If α is a root of ax2 + bx + c = 0, then we say that
(i) x = α satisfes the equation ax2 + bx + c =0
(ii) x = α is a solution of the equation ax2 + bx + c = 0
- The roots of a quadratic equation ax2 + bx + c = 0 are called the zeros of the polynomial ax2 + bx + c.
- Solving a quadratic equation means finding its roots.
- If ax2 + bx + c can be factorised as (x – α) (x – β), then ax2 + bx + c = 0 is equivalent to (x – α) (x – β) = 0
Thus, (x – α) (x – β) = 0
⇒ x – α = 0 or x – β = 0
i.e., x = α or x = β.
Here α and β are called the roots of the equation ax2 + bx + c = 0.
- To solve a quadratic equation by factorisation:
- Clear fractions and brackets, if necessary.
- Transfer all the terms to L.H.S. and combine like terms.
- Write the equation in the standard form,
i.e., ax2 + bx + c = 0.
- Factorise the L.H.S.
- Put each factor equal to zero and solve.
- Check each value by substituting it in the given equation.
- The roots of a quadratic equation can also be found by using the method of completing the square.
- The roots of the quadratic equation ax2 + bx + c = 0, a, b, c ∈ R and a ≠ 0 are given by Shridharacharya’s formula
- The expression b2 – 4ac is called the discriminant of the quadratic equation ax2 + bx + c = 0.
- The discriminant, usually denoted by D, decides the nature of roots of a quadratic equation.
- If D > 0, the equation has real roots and roots are unequal,
i.e., unequal-real roots.
If D is a perfect square, the equation has unequal-rational roots.
- If D = 0, the equation has real and equal roots and each root is .
- If D < 0, the equation has no real roots.
- If – p ≥ 5, then p ≤ – 5
- If – p ≥ –5, then p ≤ – 5
- If p2 ≥ 4, then either p ≤ –2 or p ≥ – 5
- If p2 ≤ 4, then p lies between –2 and 2,
i.e.,–2 ≤ p ≤ 2
- Quadratic equations can be applied to solve word problems involving various situations.
To solve problems leading to quadratic equations, following steps may be used :
- Represent the unknown quantity in the problem by a variable (letter).
- Translate the problem into an equation involving this variable.
- Solve the equation for the variable.
- Check the result by satisfying the conditions of the original problem.
- A root of the quadratic equation, which does not satisfy the conditions of the problem, must be rejected.
- Find the values of k for which roots of the equation x2 – 8kx + 2k = 0 are equal.
- Find the values of p such that the quadratic equation (p – 12)x2 – 2 (p – 12)x + 2 = 0 has equal roots.
- Solve the equation : 10ax2 + 15ax – 6x – 9 = 0, a ≠ 0.
- Solve for x : 4x2 – 2(a2 + b2)x + a2b2 = 0.
- For what value of k, (4 – k)x2 + (2k + 4) x + (8k + 1) = 0, is a perfect square.
- Solve for x:
- If –5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the values of p and k.
- If the roots of the equation x2 + 2cx + ab = 0 are real and unequal, prove that the equation x2 – 2(a + b)x + a2 + b2 + 2c2 = 0 has no real roots.
- If p, q, r and s are real numbers such that pr = 2(q + s), then show that at least one of the equations x2 + px + q = 0 and x2 + rx + s = 0 has real roots.
- If the equation (1 + m2)x2 + 2mcx + (c2 – a2) = 0 has equal roots, prove that c2 = a2(1 + m2).
- Prove that the equation x2(a2 + b2) + 2x(ac + bd) + (c2 + d2) = 0 has no real root, if ad ≠ bc.
- Three consecutive positive integers are taken such that the sum of the square of the first and the product of the other two is 154. Find the integers.
- The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is .
- The product of the digits of a two digit positive number is 24. If 18 is added to the number, then the digits of the number are interchanged. Find the number.
- A two digit number is four times the sum of its digits. It is also equal to three times the product of its digits. Find the number.
- The differene of squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.
- The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is , find the fraction.
- The numerator of a fraction is 3 less than its denominator. If 2 is added to both numerator as well as denominator, the sum of new and original fraction is , find the fraction.
- The hypotenuse of a right angled triangle is 6 cm more than twice its shortest side. If third side is 2 cm less than the hypotenuse, find the sides of this triangle.
- The hypotenuse of a right triangle is 3√5 cm. If the smaller side is tripled and the larger side is doubled, the new hypotenuse will be 15 cm. Find the length of each side.
- Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
- A person has a rectangular garden whose area is 100 sq m. He fences three sides of the garden with 30 m barbed wire. On the fourth side, the wall of his house is constructed; find the dimensions of the garden.
- Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
- Two years ago a man’s age was three times the square of his son’s age. Three years hence his age will be four times his son’s age. Find their present ages.
- The product of Tanay’s age (in years) five years ago and his age ten years later is 16. Determine Tanay’s present age.
- The sum of ages of father and his son is 45 years. 5 years ago, the product of their ages was 124. Determine their present ages.
- At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.
- Some students planned a picnic.The budget for food was ₹ 480. But 8 of them failed to go, the cost of food for each member increased by ₹ 10. How many students attended the picnic?
- A person on tour has ₹ 4200 for his expenses. If he extends his tour for 3 days, he has to cut down his daily expenses by ₹ 70. Find the duration of the tour.
- A trader bought a number of articles for ₹ 900, five articles were found damaged. He sold each of the remaining articles at ₹ 2 more than what he paid for it. He got a proft of ₹ 80 on the whole transaction. Find the number of articles he bought.
- A man bought a certain number of toys for ₹ 180. He kept one for his own use and sold the rest for one rupee each more than he gave for them. Besides getting his own toy for nothing, he made a profit of ₹ 10. Find the number of toys, he initially bought.
- A factory produces certain pieces of pottery in a day. It was observed on a particular day that the cost of production of each piece (in rupees) was 3 more than twice the number of articles produced in the day. If the total cost of production on that day was ₹ 90, fnd the number of pieces produced and cost of each piece.
- In a class test the sum of Gagan’s, marks in Mathematics and English is 45. If he had 1 more mark in Maths and 1 less in English, the product of marks would have been 500. Find the original marks obtained by Gagan in Maths and English separately.
- In a class test, the sum of Kamal’s marks in Maths and English is 40. Had he got 3 marks more in Maths and 4 marks less in English, the product of their marks would have been 360. Find his marks in two subjects.
- Two pipes can together fill a tank in minutes. If one pipe takes 3 minutes more than the other to fll it, find the time in which each pipe can fll the tank.
- A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4 days, find the time taken by B to finish the work.
- A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train was 10 km/hr less than that of the fast train, find the speeds of the trains.
- By increasing the speed of a bus by 10 km/hr, it takes one and half hours less to cover a journey of 450 km. Find the original speed of the bus.
- A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
- A train travels 300 km at a uniform speed. If the speed of the train had been 5 km/hour more, it would have taken 2 hours less for the same journey. Find the usual speed of the train.
- A train travels at a uniform speed for a distance of 63 km and then travels a distance of 72 km at an aveage speed of 6 km/h more than its original speed. If it takes 3 hours to complete the total journey, what is the original speed of the train?
- Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
- The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return downstream to the original point in 4 hours 30 minutes. Find the speed of the stream.
- A motorboat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of stream.
- A peacock is sitting on the top of a pillar which is 9 m high. From a point 27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake the peacock pounces on it. If their speeds are equal, at what distane from the hole is the snake caught?
- Out of a number of Saras birds, one fourth the number are moving about in lotus plants, 1/9 th coupled (along) with 1/4 as well as 7 times the square root of the number move on a hill, 56 birds remain in Vakula trees. What is the total number of birds?